Factoring x^4 + x^3 + 2x - 4 = 0 (cubic equ)

[jenettezone]

Homework Statement

x^4 + x^3 + 2x - 4 = 0

Homework Equations

N/A

The Attempt at a Solution

x^4 + x^3 + 2x - 4 = 0
x(x^3 + x^2 +2) = 4

i don't know what to do with this. i tried to factor (x^3 + x^2 +2), but i don't know how. I also have a feeling that I am not doing this correctly and that there should be a zero instead of a 4 on the right hand side of the equal sign...

 

[Borek]

This is not cubic.

Perhaps http://en.wikipedia.org/wiki/Rational_root_theorem would help (especially as a₄ = 1).

 

[Matterwave]

I don't think there is, in general, a good way to factor a 4th degree polynomial. You can try synthetic division, if you think you have one factor: (x-1).

 

[Borek]

Rational root theorem sure seems like an overkill in this case.

So what you do is you say aha! x = 1 and x = -2

1 and -2 are between obvious root candidates pointed to by the rational root theorem - so you have just used it.

Besides, you have also just solved the question for the OP, which is exactly a thing that you should not do.

 

[clamtrox]

1 and -2 are between obvious root candidates pointed to by the rational root theorem - so you have just used it.

Besides, you have also just solved the question for the OP, which is exactly a thing that you should not do.

Oops, my bad. Also, I most definitely did not use rational root theorem; I used guessing. Just because I guess something and there exists a theorem that says my guess is good, doesn't mean I know or in any way care about the theorem. :-) Still, obviously it's a nice thing to know -- I wasn't thinking at all when posting.

 

[Borek]

I wasn't thinking at all when posting.

:rofl: happens to everyone

 

[HallsofIvy]

This is not cubic.

Perhaps http://en.wikipedia.org/wiki/Rational_root_theorem would help (especially as a₄ = 1).

But the polynomial he gets after factoring out x-1 is a cubic. Perhaps that is what he was talking about.

And the rational root theorem works nicely to find a rational root of that cubic, leaving just a quadratic equation to be solved. (The quadratic has complex roots.)